What's wrong with this lasso regression implementation in R? - r

I'm trying to code from scratch a function that estimates regression coefficients using LASSO with coordinated descent (Gauss-Seidel) and soft thresholding.
My code is the following:
library(MASS)
set.seed(1)
n = 200
p = 200
V = matrix(0.2, p, p)
diag(V) = 1
X = as.matrix(mvrnorm(n, mu = rep(0, p), Sigma = V))
y = X[, 1] + 0.5*X[, 2] + 0.25*X[, 3] + rnorm(n)
X = scale(X)
y = scale(y)
soft_th <- function(b, lambda){
if (b > lambda){
return (b - lambda)
}
else if (b < -lambda){
return(b + lambda)
}
else {
return (0)
}
}
myLasso <- function(X,y, lambda=0.3,tol=1e-5,maxitr=100){
beta_old <- rep(0,p)
beta_new <- rep(0,p)
for(i in 1:maxitr){
beta_old <- beta_new
for (j in (1:ncol(X)))
{
X_j <- X[,j]
y_pred <- t(X)%*%beta_old
rho <- t(X_j)%*%(y - y_pred + beta_old[j]*X_j)
beta_new[j] <- soft_th(rho,0.7)
}
l1 <- sum(abs(beta_old-beta_new))
print(l1)
r <- y - t(X)%*%beta_old
if (l1<tol){
print('Convergence reached')
break
}
}
}
myLasso(X,y)
The problem that I have is that the L1 norm between beta_old and beta_new increases (a lot!) between each iteration. I'm following what is said in this post:
https://stats.stackexchange.com/questions/123672/coordinate-descent-soft-thresholding-update-operator-for-lasso/351134#351134
I think that somewhat I'm not correctly implementing the descent update rule.
Any help would be appreciated. Thanks in advance.

I was doing more research, and it seems that I didn't normalise the X matrix. After adding X <- X/norm(X,type='2') after defining X, the problem solved.
Now a new problem that I have is that this new function does not replicate the results of the glmnet implementation of LASSO regression. What could it be? I got a RMSE of 0.6 with glmnet and 0.997 with my implementation. I'd love if someone could guide me in how to improve my function.
Thanks in advance.

Related

Derivative of a function with matrices and vectors for Newton-Raphson method

I've tried to find roots of nonlinear equation by using a Newton-Raphson method.
The problem I stuck in is that I don't get the right first derivative of following equation:
y is the target variable for my prediction, X is a matrix of the predictor variables, Theta is a smoothing parameter.
I need to get the arg min of Q.
For that I want to use this approach of Newton-Raphson:
newton.raphson <- function(f, a, b, tol = 1e-5, n = 1000) {
require(numDeriv) # Package for computing f'(x)
x0 <- a # Set start value to supplied lower bound
k <- n # Initialize for iteration results
# Check the upper and lower bounds to see if approximations result in 0
fa <- f(a)
if (fa == 0.0) {
return(a)
}
fb <- f(b)
if (fb == 0.0) {
return(b)
}
for (i in 1:n) {
dx <- genD(func = f, x = x0)$D[1] # First-order derivative f'(x0)
x1 <- x0 - (f(x0) / dx) # Calculate next value x1
k[i] <- x1 # Store x1
if (abs(x1 - x0) < tol) {
root.approx <- tail(k, n=1)
res <- list('root approximation' = root.approx, 'iterations' = k)
return(res)
}
# If Newton-Raphson has not yet reached convergence set x1 as x0 and continue
x0 <- x1
}
print('Too many iterations in method')
Thanks in advance for your help!

Sequential Quadratic Programming in R to find optimal weights of an Equally-Weighted Risk Contribution Portfolio

Introduction to the problem
I am trying to write down a code in R so to obtain the weights of an Equally-Weighted Contribution (ERC) Portfolio. As some of you may know, the portfolio construction was presented by Maillard, Roncalli and Teiletche.
Skipping technicalities, in order to find the optimal weights of an ERC portfolio one needs to solve the following Sequential Quadratic Programming problem:
with:
Suppose we are analysing N assets. In the above formulas, we have that x is a (N x 1) vector of portfolio weights and Σ is the (N x N) variance-covariance matrix of asset returns.
What I have done so far
Using the function slsqp of the package nloptr which solves SQP problems, I would like to solve the above minimisation problem. Here is my code. Firstly, the objective function to be minimised:
ObjFuncERC <- function (x, Sigma) {
sum <- 0
R <- Sigma %*% x
for (i in 1:N) {
for (j in 1:N) {
sum <- sum + (x[i]*R[i] - x[j]*R[j])^2
}
}
}
Secondly, the starting point (we start by an equally-weighted portfolio):
x0 <- matrix(1/N, nrow = N, ncol = 1)
Then, the equality constraint (weights must sum to one, that is: sum of the weights minus one equal zero):
heqERC <- function (x) {
h <- numeric(1)
h[1] <- (t(matrix(1, nrow = N, ncol = 1)) %*% x) - 1
return(h)
}
Finally, the lower and upper bounds constraints (weights cannot exceed one and cannot be lower than zero):
lowerERC <- matrix(0, nrow = N, ncol = 1)
upperERC <- matrix(1, nrow = N, ncol = 1)
So that the function which should output optimal weights is:
slsqp(x0 = x0, fn = ObjFuncERC, Sigma = Sigma, lower = lowerERC, upper = upperERC, heq = heqERC)
Unfortunately, I do not know how to share with you my variance-covariance matrix (which takes name Sigma and is a (29 x 29) matrix, so that N = 29) so to reproduce my result, still you can simulate one.
The output error
Running the above code yields the following error:
Error in nl.grad(x, fn) :
Function 'f' must be a univariate function of 2 variables.
I have no idea what to do guys. Probably, I have misunderstood how things must be written down in order for the function slsqp to understand what to do. Can someone help me understand how to fix the problem and get the result I want?
UPDATE ONE: as pointed out by #jogo in the comments, I have updated the code, but it still produces an error. The code and the error above are now updated.
UPDATE 2: as requested by #jaySf, here is the full code that allows you to reproduce my error.
## ERC Portfolio Test
# Preliminary Operations
rm(list=ls())
require(quantmod)
require(nloptr)
# Load Stock Data in R through Yahoo! Finance
stockData <- new.env()
start <- as.Date('2014-12-31')
end <- as.Date('2017-12-31')
tickers <-c('AAPL','AXP','BA','CAT','CSCO','CVX','DIS','GE','GS','HD','IBM','INTC','JNJ','JPM','KO','MCD','MMM','MRK','MSFT','NKE','PFE','PG','TRV','UNH','UTX','V','VZ','WMT','XOM')
getSymbols.yahoo(tickers, env = stockData, from = start, to = end, periodicity = 'monthly')
# Create a matrix containing the price of all assets
prices <- do.call(cbind,eapply(stockData, Op))
prices <- prices[-1, order(colnames(prices))]
colnames(prices) <- tickers
# Compute Returns
returns <- diff(prices)/lag(prices)[-1,]
# Compute variance-covariance matrix
Sigma <- var(returns)
N <- 29
# Set up the minimization problem
ObjFuncERC <- function (x, Sigma) {
sum <- 0
R <- Sigma %*% x
for (i in 1:N) {
for (j in 1:N) {
sum <- sum + (x[i]*R[i] - x[j]*R[j])^2
}
}
}
x0 <- matrix(1/N, nrow = N, ncol = 1)
heqERC <- function (x) {
h <- numeric(1)
h[1] <- t(matrix(1, nrow = N, ncol = 1)) %*% x - 1
}
lowerERC <- matrix(0, nrow = N, ncol = 1)
upperERC <- matrix(1, nrow = N, ncol = 1)
slsqp(x0 = x0, fn = ObjFuncERC, Sigma = Sigma, lower = lowerERC, upper = upperERC, heq = heqERC)
I spotted several mistakes in your code. For instance, ObjFuncERC is not returning any value. You should use the following instead:
# Set up the minimization problem
ObjFuncERC <- function (x, Sigma) {
sum <- 0
R <- Sigma %*% x
for (i in 1:N) {
for (j in 1:N) {
sum <- sum + (x[i]*R[i] - x[j]*R[j])^2
}
}
sum
}
heqERC doesn't return anything too, I also changed your function a bit
heqERC <- function (x) {
sum(x) - 1
}
I made those changes and tried slsqp without lower and upper and it worked. Still, another thing to consider is that you set lowerERC and upperERC as matrices. Use the following instead:
lowerERC <- rep(0,N)
upperERC <- rep(1,N)
Hope this helps.

How do I find the maximum likelihood of a specific multivariate normal log likelihood in R?

I'm having trouble optimizing a multivariate normal log-likelihood in R. If anyone has a good solution for that, please let me know. Specifically, I cannot seem to keep the variance-covariance matrix positive-definite and the parameters in a reasonable range.
Let me introduce the problem more completely. I am essentially trying to simultaneously solve these two regression equations using MLE:
$$
y_1 = \beta_1 + \beta_2 x_1 + \beta_3 x_2 \\
y_2 = \beta_4 + \beta_3 x_1 + \beta_5 x_2
$$
The fact that $\beta_3$ is in both equations is not a mistake. I try to solve this using MLE by maximizing the likelihood of the multivariate normal distribution for $Y = (y_1, y_2)^\top$ where the mean is parameterized as above in the regression equations.
I've attached the log-likelihood function as I believe it should be, where I constrain the variance covariance matrix to be positive-definite by recreating it from necessarily positive eigenvalues and a cholesky decomposition.
mvrestricted_ll <- function(par, Y, X) {
# Indices
n <- nrow(X)
nbetas <- (2 + 3 * (ncol(Y) - 1))
# Extract parameters
beta <- par[1:nbetas]
eigvals <- exp(par[(nbetas + 1):(nbetas + ncol(Y))]) # constrain to be positive
chole <- par[(nbetas + ncol(Y) + 1):(nbetas + ncol(Y) + ncol(Y)*(ncol(Y)+1)/2)]
# Build Sigma from positive eigenvalues and cholesky (should be pos def)
L <- diag(ncol(Y))
L[lower.tri(L, diag=T)] <- chole
Sigma <- diag(eigvals) + tcrossprod(L)
# Linear predictor
# Hard coded for 2x2 example for now
mu <- cbind(beta[1] + beta[2]*X[,1] + beta[3]*X[,2],
beta[4] + beta[3]*X[,1] + beta[5]*X[,2])
yminmu <- Y - mu
nlogs <- n * log(det(Sigma))
invSigma <- solve(Sigma)
meat <- yminmu %*% tcrossprod(invSigma, yminmu)
return(- nlogs - sum(diag(meat)))
}
# Create fake data
n <- 1000
p <- 2
set.seed(20160201)
X <- matrix(rnorm(n*p), nrow = n)
set.seed(20160201)
Y <- matrix(rnorm(n*p), nrow = n)
# Initialize parameters
initpars <- c(rep(0, (2 + 3 * (ncol(Y) - 1)) + ncol(Y) + ncol(Y)*(ncol(Y)+1)/2))
# Optimize fails with BFGS
optim(par = initpars, fn = mvrestricted_ll, X=X, Y=Y, method = "BFGS")
# Optim does not converge with Nelder-mead, if you up the maxits it also fails
optim(par = initpars, fn = mvrestricted_ll, X=X, Y=Y)
Any help would be greatly appreciated.
EDIT: I should note that just letting Sigma be a vector in the parameters and then returning a very large value whenever it is not positive definite does not work either.
I have no idea if the code/answer is correct, but
invSigma <- try(solve(Sigma))
if (inherits(invSigma, "try-error")) return(NA)
and running
optim(par = initpars, fn = mvrestricted_ll, X=X, Y=Y,
control = list(maxit = 1e5))
gets me a little farther to a convergence code of 10 (degenerate Nelder-Mead simplex).
$par
[1] 1.361612e+01 4.674349e+01 -3.050170e+01 3.305013e+01 6.731194e+01
[6] -3.117192e+01 -5.408598e+00 -6.326897e-07 -1.987449e+01 -1.795924e+01
$value
[1] -1.529013e+19
$counts
function gradient
1219 NA
$convergence
[1] 10
I suspect that a real solution will involve looking more carefully at the code to see if it's really doing what you think it's doing (sorry); understanding why solve() errors occur might be a good first step. You can work on troubleshooting this by putting a cat(par, "\n") as the first line of the function and running it without the try/NA-return code. That will allow you to isolate an example data set that throws the error — then you can work your way through your code a line at a time (with debug() or by hand) to see what's happening.
You can consider using the following approach :
library(DEoptim)
fn <- function(par, mat_X, mat_Y)
{
X <- mat_X
Y <- mat_Y
n <- nrow(X)
nbetas <- (2 + 3 * (ncol(Y) - 1))
beta <- par[1 : nbetas]
eigvals <- exp(par[(nbetas + 1) : (nbetas + ncol(Y))])
chole <- par[(nbetas + ncol(Y) + 1) : (nbetas + ncol(Y) + ncol(Y) * (ncol(Y) + 1) / 2)]
L <- diag(ncol(Y))
L[lower.tri(L, diag = TRUE)] <- chole
Sigma <- tryCatch(diag(eigvals) + tcrossprod(L), error = function(e) NA)
if(is.null(dim(Sigma)))
{
return(10 ^ 30)
}else
{
mu <- cbind(beta[1] + beta[2] * X[,1] + beta[3] * X[,2],
beta[4] + beta[3] * X[,1] + beta[5] * X[,2])
yminmu <- Y - mu
nlogs <- n * log(det(Sigma))
invSigma <- tryCatch(solve(Sigma), error = function(e) NA)
if(is.null(dim(invSigma)))
{
return(10 ^ 30)
}else
{
meat <- yminmu %*% tcrossprod(invSigma, yminmu)
log_Lik <- - nlogs - sum(diag(meat))
if(is.na(log_Lik) | is.nan(log_Lik) | is.infinite(log_Lik))
{
return(10 ^ 30)
}else
{
return(-log_Lik)
}
}
}
}
n <- 1000
p <- 2
set.seed(20160201)
mat_X <- matrix(rnorm(n * p), nrow = n)
set.seed(2436537)
mat_Y <- matrix(rnorm(n * p), nrow = n)
lower <- rep(-10, 10)
upper <- rep(10, 10)
DEoptim(fn = fn, lower = lower, upper = upper,
control = list(itermax = 10000, parallelType = 1), mat_X = mat_X, mat_Y = mat_Y)

How to find the second derivative in R and while using newton's method with numerical derivation

The log-likelihood of the gamma distribution with scale parameter 1 can be written as:
(α−1)s−nlogΓ(α)
where alpha is the shape parameter and s=∑logXi is the sufficient statistic.
Randomly draw a sample of n = 30 with a shape parameter of alpha = 4.5. Using newton_search and make_derivative, find the maximum likelihood estimate of alpha. Use the moment estimator of alpha, i.e., mean of x as the initial guess. The log-likelihood function in R is:
x <- rgamma(n=30, shape=4.5)
gllik <- function() {
s <- sum(log(x))
n <- length(x)
function(a) {
(a - 1) * s - n * lgamma(a)
}
}
I have created the make_derivative function as follows:
make_derivative <- function(f, h) {
(f(x + h) - f(x - h)) / (2*h)
}
I also have created a newton_search function that incorporates the make_derivative function; However, I need to use newton_search on the second derivative of the log-likelihood function and I'm not sure how to fix the following code in order for it to do that:
newton_search2 <- function(f, h, guess, conv=0.001) {
set.seed(2)
y0 <- guess
N = 1000
i <- 1; y1 <- y0
p <- numeric(N)
while (i <= N) {
make_derivative <- function(f, h) {
(f(y0 + h) - f(y0 - h)) / (2*h)
}
y1 <- (y0 - (f(y0)/make_derivative(f, h)))
p[i] <- y1
i <- i + 1
if (abs(y1 - y0) < conv) break
y0 <- y1
}
return (p[(i-1)])
}
Hint: You must apply newton_search to the first and second derivatives (derived numerically using make_derivative) of the log-likelihood. Your answer should be near 4.5.
when I run newton_search2(gllik(), 0.0001, mean(x), conv = 0.001), I get double what the answer should be.
I re-wrote the code and it works perfectly now (even better than what I had originally wrote). Thanks to all who helped. :-)
newton_search <- function(f, df, guess, conv=0.001) {
set.seed(1)
y0 <- guess
N = 100
i <- 1; y1 <- y0
p <- numeric(N)
while (i <= N) {
y1 <- (y0 - (f(y0)/df(y0)))
p[i] <- y1
i <- i + 1
if (abs(y1 - y0) < conv) break
y0 <- y1
}
return (p[(i-1)])
}
make_derivative <- function(f, h) {
function(x){(f(x + h) - f(x - h)) / (2*h)
}
}
df1 <- make_derivative(gllik(), 0.0001)
df2 <- make_derivative(df1, 0.0001)
newton_search(df1, df2, mean(x), conv = 0.001)

Newton Raphson for logistic regression

I did code for Newton Raphson for logistic regression. Unfortunately I tried many data there is no convergence. there is a mistake I do not know where is it. Can anyone help to figure out what is the problem.
First the data is as following; y indicate the response (0,1) , Z is 115*30 matrix which is the exploratory variables. I need to estimate the 30 parameters.
y = c(rep(0,60),rep(1,55))
X = sample(c(0,1),size=3450,replace=T)
Z = t(matrix(X,ncol=115))
#The code is ;
B = matrix(rep(0,30*10),ncol=10)
B[,1] = matrix(rep(0,30),ncol=1)
for(i in 2 : 10){
print(i)
p <- exp(Z %*%as.matrix(B[,i])) / (1 + exp(Z %*% as.matrix(B[,i])))
v.2 <- diag(as.vector(1 * p*(1-p)))
score.2 <- t(Z) %*% (y - p) # score function
increm <- solve(t(Z) %*% v.2 %*% Z)
B[,i] = as.matrix(B[,i-1])+increm%*%score.2
if(B[,i]-B[i-1]==matrix(rep(0.0001,30),ncol=1)){
return(B)
}
}
Found it! You're updating p based on B[,i], you should be using B[,i-1] ...
While I was finding the answer, I cleaned up your code and incorporated the results in a function. R's built-in glm seems to work (see below). One note is that this approach is likely to be unstable: fitting a binary model with 30 predictors and only 115 binary responses, and without any penalization or shrinkage, is extremely optimistic ...
set.seed(101)
n.obs <- 115
n.zero <- 60
n.pred <- 30
y <- c(rep(0,n.zero),rep(1,n.obs-n.zero))
X <- sample(c(0,1),size=n.pred*n.obs,replace=TRUE)
Z <- t(matrix(X,ncol=n.obs))
R's built-in glm fitter does work (it uses iteratively reweighted least squares, not N-R):
g1 <- glm(y~.-1,data.frame(y,Z),family="binomial")
(If you want to view the results, library("arm"); coefplot(g1).)
## B_{m+1} = B_m + (X^T V_m X)^{-1} X^T (Y-P_m)
NRfit function:
NRfit <- function(y,X,start,n.iter=100,tol=1e-4,verbose=TRUE) {
## used X rather than Z just because it's more standard notation
n.pred <- ncol(X)
B <- matrix(NA,ncol=n.iter,
nrow=n.pred)
B[,1] <- start
for (i in 2:n.iter) {
if (verbose) cat(i,"\n")
p <- plogis(X %*% B[,i-1])
v.2 <- diag(c(p*(1-p)))
score.2 <- t(X) %*% (y - p) # score function
increm <- solve(t(X) %*% v.2 %*% X)
B[,i] <- B[,i-1]+increm%*%score.2
if (all(abs(B[,i]-B[,i-1]) < tol)) return(B)
}
B
}
matplot(res1 <- t(NRfit(y,Z,start=coef(g1))))
matplot(res2 <- t(NRfit(y,Z,start=rep(0,ncol(Z)))))
all.equal(res2[6,],unname(coef(g1))) ## TRUE

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